It would be one thing to just connect a ground somewhere, but that will ruin the validity of my circuit.

The voltage source defined as MAINS in this circuit is just that...a simulation of the mains. It is a 60Hz sine wave with an amplitude of 120V. If I add a ground to it, I won't get the negative part of my wave!

The resistor connected to the voltage source is there to prevent another error being thrown about having inductors in parallel with the voltage source.

How can I keep this circuit equivalent to the real world so that I can simulate without losing half my wave from my MAINS voltage source?

If your local mains supply is say stated as 120V 60Hz, it means its 120V RMS.
The actual peak voltage of the 120V RMS is 1.414 *120V = ~169V.

This is the peak voltage applied to the input to the full wave rectifier diode bridge, which means the capacitor C1 will charge to close to 169V DC [ there are diode voltage drops, say 1.4V total] so thats around ~167.5V !!!

If your local mains supply is say stated as 120V 60Hz, it means its 120V RMS.
The actual peak voltage of the 120V RMS is 1.414 *120V = ~169V.

This is the peak voltage applied to the input to the full wave rectifier diode bridge, which means the capacitor C1 will charge to close to 169V DC [ there are diode voltage drops, say 1.4V total] so thats around ~167.5V !!!

This will of course be applied to the Load via the circuit.

OK.

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That's definitely good to know. It will effect my transformer. But, that's all that will change...the transformer. At least I think?

Again, 12V would be considered a RMS value, so the peak would be approx 17V [ less two diode drops after the rectifier] ie: ~ 15.5Vdc

I would use 10H for the transformer primary and 100mH for the secondary, thats a 10:1 step down, also set the voltage to 169V for the mains.

Is the load only a 10K resistance.?

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No, that's just a dummy load. The resistance of my control electronics will be changing that's why I went with the buck converter. Let me see if I can come up with a formula to get the peak voltage I want. Should be just some Algebra.

1.4142 is the square root of 2, often written as sqrt(2).
That is the ratio of the Peak voltage to RMS voltage of a sine wave.
So if your RMS voltage is 120v, the peak is 120*1.4142=169.7 volts, which is the highest point on the sine wave.
A more accurate calculation is:
sqrt(2)=1.4142135623730950488016887242097
120*sqrt(2)=169.70562748477140585620264690516

If you already know the peak value and you want to know what the RMS value is, just divide by the sqrt(2).